Shortest Path in DAG Using Topological Order in DSA C# - Time & Space Complexity

Time Complexity: Shortest Path in DAG Using Topological Order

O(V + E)

Understanding Time Complexity

We want to understand how the time needed to find the shortest path in a Directed Acyclic Graph (DAG) changes as the graph grows.

Specifically, how does the algorithm's work increase when the number of nodes and edges increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Find shortest paths in DAG using topological order
void ShortestPathDAG(int V, List<(int, int)>[] graph, int src) {
    int[] dist = new int[V];
    for (int i = 0; i < V; i++) dist[i] = int.MaxValue;
    dist[src] = 0;

    var topoOrder = TopologicalSort(V, graph);

    foreach (var u in topoOrder) {
        if (dist[u] != int.MaxValue) {
            foreach (var (v, w) in graph[u]) {
                if (dist[u] + w < dist[v]) dist[v] = dist[u] + w;
            }
        }
    }
}
    

This code finds shortest paths from a source node to all others in a DAG by first sorting nodes topologically, then relaxing edges in order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping over all nodes and their outgoing edges to update distances.
• How many times: Each node is visited once in topological order, and each edge is checked once during relaxation.

How Execution Grows With Input

As the number of nodes (V) and edges (E) grow, the algorithm visits each node once and checks each edge once.

Input Size (V, E)	Approx. Operations
10 nodes, 15 edges	~25 operations (10 + 15)
100 nodes, 200 edges	~300 operations (100 + 200)
1000 nodes, 5000 edges	~6000 operations (1000 + 5000)

Pattern observation: The work grows roughly in proportion to the sum of nodes and edges.

Final Time Complexity

Time Complexity: O(V + E)

This means the time needed grows linearly with the number of nodes plus the number of edges in the graph.

Common Mistake

[X] Wrong: "The algorithm checks every possible path, so it must be exponential time."

[OK] Correct: Because the graph is acyclic and we use topological order, each edge is relaxed only once, avoiding repeated work on paths.

Interview Connect

Understanding this time complexity shows you can analyze graph algorithms that smartly avoid repeated work by using orderings like topological sort.

This skill helps you explain why some graph problems can be solved efficiently despite many possible paths.

Self-Check

"What if the graph had cycles? How would the time complexity and approach change?"

Practice

1. What is the main advantage of using topological order to find the shortest path in a Directed Acyclic Graph (DAG)?

easy

A. It sorts edges by weight.

B. It guarantees the graph has cycles.

C. It allows processing vertices in a linear order without revisiting nodes.

D. It finds the longest path instead of the shortest.

Solution

1. Step 1: Understand topological order in DAG

   Topological order arranges vertices so that all edges go from earlier to later vertices, ensuring no cycles.

2. Step 2: Use topological order for shortest path

   Processing vertices in this order allows updating shortest distances without revisiting nodes, making the algorithm efficient.

3. Final Answer:

   It allows processing vertices in a linear order without revisiting nodes. -> Option C

4. Quick Check:

   Topological order = linear processing [OK]

Hint: Topological order processes nodes once in DAG [OK]

Common Mistakes:

• Thinking topological order works with cycles
• Confusing shortest path with longest path
• Assuming edges are sorted by weight

2. Which of the following C# code snippets correctly initializes the distance array for shortest path calculation in a DAG with 5 vertices?

easy

A. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue;

B. int[] dist = new int[5] {0, 0, 0, 0, 0};

C. int[] dist = new int[5]; dist.Fill(int.MaxValue);

D. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = 0;

Solution

1. Step 1: Understand distance initialization

   Distances should start as infinity (int.MaxValue) except the source, to represent unknown paths.

2. Step 2: Check code correctness

   int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue; correctly sets all distances to int.MaxValue using a loop. int[] dist = new int[5] {0, 0, 0, 0, 0}; sets all to zero, which is incorrect. int[] dist = new int[5]; dist.Fill(int.MaxValue); uses a non-existent Fill method. int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = 0; sets all to zero, which is wrong.

3. Final Answer:

   int[] dist = new int[5]; for(int i = 0; i < 5; i++) dist[i] = int.MaxValue; -> Option A

4. Quick Check:

   Initialize distances to max value [OK]

Hint: Initialize distances with int.MaxValue except source [OK]

Common Mistakes:

• Setting all distances to zero
• Using non-existent array methods
• Not initializing distances before algorithm

3. Consider the following DAG with edges and weights:
0 -> 1 (2), 0 -> 2 (4), 1 -> 2 (1), 1 -> 3 (7), 2 -> 4 (3), 3 -> 4 (1)
If we run the shortest path algorithm from vertex 0 using topological order, what is the shortest distance to vertex 4?

medium

A. 10

B. 6

C. 7

D. 9

Solution

1. Step 1: Find topological order and initialize distances

   Topological order can be [0,1,2,3,4]. Initialize dist[0]=0, others = infinity.

2. Step 2: Relax edges in topological order

   From 0: dist[1]=2, dist[2]=4.
   From 1: dist[2]=min(4,2+1=3)=3, dist[3]=2+7=9.
   From 2: dist[4]=3+3=6.
   From 3: dist[4]=min(6,9+1=10)=6.

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Shortest distance to 4 = 6 [OK]

Hint: Relax edges in topological order to update distances [OK]

Common Mistakes:

• Not updating distances after relaxation
• Using wrong topological order
• Ignoring better paths through intermediate nodes

4. The following C# code attempts to find shortest paths in a DAG but has a bug:

int[] dist = new int[V];
for(int i = 0; i < V; i++) dist[i] = int.MaxValue;
dist[source] = 0;
foreach(int u in topologicalOrder)
{
  foreach(var edge in graph[u])
  {
    int v = edge.Item1;
    int weight = edge.Item2;
    if(dist[u] + weight < dist[v])
      dist[v] = dist[u] + weight;
  }
}

What is the bug in this code?

medium

A. No check if dist[u] is int.MaxValue before adding weight

B. Edges are not relaxed in correct order

C. topologicalOrder is not computed

D. dist array is not initialized to int.MaxValue

Solution

1. Step 1: Analyze distance update condition

   The code adds dist[u] + weight without checking if dist[u] is int.MaxValue, which can cause overflow.

2. Step 2: Identify missing check

   Before adding weight, code should check if dist[u] != int.MaxValue to avoid invalid calculations.

3. Final Answer:

   No check if dist[u] is int.MaxValue before adding weight -> Option A

4. Quick Check:

   Check dist[u] before adding weight [OK]

Hint: Check dist[u] != int.MaxValue before adding weight [OK]

Common Mistakes:

• Ignoring overflow from adding to int.MaxValue
• Assuming topological order is always computed
• Thinking initialization is missing

5. Given a DAG with vertices 0 to 5 and edges:
0->1(3), 0->2(6), 1->2(4), 1->3(4), 1->4(11), 2->3(8), 3->4(-4), 4->5(2)
Using shortest path with topological order from vertex 0, what is the shortest distance to vertex 5?

hard

A. 10

B. 9

C. 7

D. 5

Solution

1. Step 1: Determine topological order and initialize distances

   Possible topological order: [0,1,2,3,4,5]. Initialize dist[0]=0, others = infinity.

2. Step 2: Relax edges in order

   From 0: dist[1]=3, dist[2]=6.
   From 1: dist[2]=min(6,3+4=7)=6, dist[3]=3+4=7, dist[4]=3+11=14.
   From 2: dist[3]=min(7,6+8=14)=7.
   From 3: dist[4]=min(14,7+(-4)=3)=3.
   From 4: dist[5]=3+2=5.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Shortest distance to 5 = 5 [OK]

Hint: Relax edges in topological order, watch negative weights carefully [OK]

Common Mistakes:

• Ignoring negative edge weights
• Not updating distances after better path found
• Wrong topological order causing incorrect results